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The question:Let $p, q$ and $r$ be prime other than $3.$ Show that $3$ divides $p^2+q^2+r^2.$

I am not sure how to start.

Should I use" Fundamental Theorem of Arithmetic"? I know every integer greater than $1$ is a prime or a product of primes, but I don't know how to apply on this question

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    $\begingroup$ Primes is not the point. It is just a way to make sure none of $p,q,r$ is divisible by $3$. It will be true any time none of them are divisible by $3$ (or if they all are). $\endgroup$ – Ross Millikan Sep 6 '13 at 3:06
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Hint: Work modulo $3$. All primes except $3$ are congruent to either $1 \pmod 3$ or $-1 \pmod 3$.

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  • $\begingroup$ Equivalently, assume that $p=3\hat p\pm1$, $q=3\hat q\pm1$, and $r=3\hat r\pm1$, if you haven't learned modular arithmetic. $\endgroup$ – Glen O Sep 6 '13 at 2:22
  • $\begingroup$ @Glen O $11 = 3\times 3 + {\Huge 2}$. $\endgroup$ – Felix Marin Sep 6 '13 at 2:44
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    $\begingroup$ @FelixMarin: $11 = 3\times4 - {\Huge 1}$ $\endgroup$ – Glen O Sep 6 '13 at 2:49
  • $\begingroup$ @GlenO I got it. Thanks. $\endgroup$ – Felix Marin Sep 6 '13 at 3:08
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    $\begingroup$ But 3 is not a divisor of p,q,r. In fact, the opposite is true; 3 is never a divisor of p, q nor r. Try this: "If $p \equiv 1 \pmod 3$, then $p^2 \equiv ??? \pmod 3$. If $p \equiv -1 \pmod 3$ then $p^2 \equiv ??? \pmod 3$. The same is true for $q$ and $r$. Hence $p^2+q^2+r^2 \equiv ??? \pmod 3$. $\endgroup$ – Rebecca J. Stones Sep 6 '13 at 13:54
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Given a prime number $x_{i}$ other than $3$, it can be written as $x_{i} = 3n_{i} + r_{i}$ where $n_{i}$ and $r_{i}$ are integers and $1 \leq r_{i} \leq 2$. Then

$$ \sum_{i = 1}^{3}x_{i}^{2} = 3\left(3\sum_{i = 1}^{3}n_{i}^{2} + 2\sum_{i = 1}^{3}n_{i}r_{i}\right) + \sum_{i = 1}^{3}\overbrace{\quad r_{i}^{2}\quad}^{\displaystyle{3r_{i} - 2}} = 3\left\lbrack% 3\sum_{i = 1}^{3}n_{i}^{2} + 2\sum_{i = 1}^{3}n_{i}r_{i} + \sum_{i = 1}^{3}r_{i} - 2 \right\rbrack $$

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  • $\begingroup$ Why complicating it unnecessarily? $\endgroup$ – punctured dusk Sep 6 '13 at 11:36

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